首页 > read-only variable is not assignable问题

read-only variable is not assignable问题

不知道该如何叙述这个问题,请看代码:

#include <set>
#include <iostream>
using namespace std;

struct test {
    int data[3];
};

int main(void) {
    test t1;
    set<test> s;
    s.insert(t1);
    s.begin()->data[0] = 1;//此处赋值报错:read-only variable is not assignable

    return 0;
}

应该是set内部的问题吧?
求高人解释,非常感谢...


set的迭代器是只读的,不允许修改(因为set内部有将元素排序,随意更改元素的值可能会破坏其有序性);

要改变元素的值,只能先把旧的删除,再把新的插入进去


set内的元素不可以更改,只能增加或者删除。http://www.cplusplus.com/refe...

Sets are containers that store unique elements following a specific order.

In a set, the value of an element also identifies it (the value is itself the key, of type T), and each value must be unique. The value of the elements in a set cannot be modified once in the container (the elements are always const), but they can be inserted or removed from the container.

Internally, the elements in a set are always sorted following a specific strict weak ordering criterion indicated by its internal comparison object (of type Compare).

set containers are generally slower than unordered_set containers to access individual elements by their key, but they allow the direct iteration on subsets based on their order.

Sets are typically implemented as binary search trees.

意思好像是:只读的变量是不能被赋值的。。。

【热门文章】
【热门文章】